Relativistic centripetal force

[Dale]

you need to start from the Lorentz transforms for rotating frames.

I have never heard of the Lorentz transform for rotating frame. AFAIK the Lorentz transform transforms between different inertial frames, not between an inertial frame and a rotating frame.

 

[Dale]

No, it doesn't mean that, it only means that the comoving frame (for which you are calculating the proper acceleration) is characterized by \omega=0 . Because the frame is co-moving with the rotating object.
So, you should have obtained the proper acceleration magnitude as:

a=r\omega^2

The Minkowski norm of the four-acceleration is frame invariant. So it does not matter which frame you calculate it in; it will be the same in all frames. If those factors of 1-r^2\omega^2/c^2 appear in one frame they must appear in all frames.

 

[starthaus]

OK, if I understand correctly, because of the t' contained in the cos and sin functions in the last terms, the expressions for (6) should be:

\frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\cos(\gamma\omega t')-\gamma\omega\sin(\gamma\omega t') \frac{dx'}{dt'}-\gamma\frac{d^2y'}{dt'^2}\sin(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\cos(\gamma\omega t') - R\gamma^2\omega^2\cos(\gamma\omega t')\right) (Eq6x)

\frac{d^2y}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\sin(\gamma\omega t')+\gamma\omega\cos(\gamma\omega t') \frac{dx'}{dt'}+\gamma\frac{d^2y'}{dt'^2}\cos(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\sin(\gamma\omega t') - R\gamma^2\omega^2\sin(\gamma\omega t')\right) (Eq6y)

Is that correct?

Correct. Finally.

If they are correct, then when t' = 0, \sin(\gamma\omega t') = 0 and \cos(\gamma\omega t') = 1 and the equations simplify to:

Incorrect. In the comoving frame you need to realize that \frac{dx'}{dt'}=0 and \frac{dy'}{dt'}=0. Try continuing the calculations from this hint.

 

[starthaus]

I have never heard of the Lorentz transform for rotating frame. AFAIK the Lorentz transform transforms between different inertial frames, not between an inertial frame and a rotating frame.

I gave a couple of references earlier in the thread. Here they are again:

1. R. A. Nelson, "Generalized Lorentz transformation for an accelerated, rotating frame of reference", J. Math. Phys. 28, 2379 (1987);

2. H.Nikolic, “Relativistic contraction and related effects in non-inertial frames”, Phys.Rev. A, 61, (2000)

 

[starthaus]

The Minkowski norm of the four-acceleration is frame invariant. So it does not matter which frame you calculate it in; it will be the same in all frames. If those factors of 1-r^2\omega^2/c^2 appear in one frame they must appear in all frames.

No argument. The argument is about the fact that proper acceleration is DIFFERENT from four-acceleration. It differs exactly by the \gamma factor. What you calculated iis the four-acceleration, not the proper acceleration. But, we are digressing. The discussion is about the correct usage of Lorentz transforms. Please be patient, kev is one step away from finishing the computations.

 

[Jorrie]

... In the comoving frame you need to realize that \frac{dx'}{dt'}=0 and
\frac{dy'}{dt'}=0.

It appears to me as if you have a different definition of comoving frame than the rest of us. We are usually referring to the momentarily comoving inertial frame (MCIF). Or do I read you incorrectly?

 

[starthaus]

It appears to me as if you have a different definition of comoving frame than the rest of us. We are usually referring to the momentarily comoving inertial frame (MCIF).

The comoving frame (S') is moving with the particle, hence \frac{dx'}{dt'}=\frac{dy'}{dt'}=0.

Or do I read you incorrectly?

Apparently you did read it incorrectly. Hopefully the above explanation corrected your view.

 

[Dale]

The argument is about the fact that proper acceleration is DIFFERENT from four-acceleration. It differs exactly by the \gamma factor.

Well, yes, the proper acceleration is a 3-vector and the four-acceleration is a 4-vector, so obviously they are different. But the Minkowski norm of the four-acceleration is equal to the Euclidean norm of the proper acceleration, they do not differ by a factor of \gamma. I.e. the Minkowski norm of the four-acceleration is the magnitude of the acceleration measured by an accelerometer.

P.S. I looked for both of those references. I didn't have access to the first, and I couldn't find the second at all. I can use the usual non-generalized Lorentz transform to answer any questions that can be expressed in terms of a MCIF, but I won't be able to use these non-inertial generalizations.

 

[starthaus]

I.e. the Minkowski norm of the four-acceleration is the magnitude of the acceleration measured by an accelerometer.

The wiki page contradicts you. But, we are digressing, could you please go all the way back to post #3 to get the full scope of the disagreement? It is about the proper usage of Lorentz transforms.

 

[starthaus]

P.S. I looked for both of those references. I didn't have access to the first, and I couldn't find the second at all.

I also gave the arxiv link for the second paper. The author is a very respected physicist (and a mentor on this forum, see here ).

I can use the usual non-generalized Lorentz transform to answer any questions that can be expressed in terms of a MCIF, but I won't be able to use these non-inertial generalizations.

If you are patient, kev is only one step away from finishing the calculatons using the correct "generalizations". Why not be patient and learn something new? Let's give him a little more time, I am quite sure that using the latest hint he'll come up with the correct transformations at his next iteration. He's converging on the correct derivation.

 

[Dale]

The wiki page contradicts you.

I think you must be misunderstanding something. The wikipedia page clearly says that "the magnitude of the four-acceleration (which is an invariant scalar) is equal to the proper acceleration". Although technically it should say "magnitude of the proper acceleration" since the proper acceleration is a vector.

http://en.wikipedia.org/wiki/Four-acceleration

 

[starthaus]

I think you must be misunderstanding something. The wikipedia page clearly says that "the magnitude of the four-acceleration (which is an invariant scalar) is equal to the proper acceleration". Although technically it should say "magnitude of the proper acceleration" since the proper acceleration is a vector.

http://en.wikipedia.org/wiki/Four-acceleration

...for \gamma=1. A very important detail.
Look at it this way: your formula produces an infinite acceleration for \omega->\frac{c}{r}. This is not only unphysical but also diverges from the Newtonian prediction r\omega^2=\frac{c^2}{r}. So, your application of the method for calculating 4-acceleration does not recover the finite Newtonian predictions in the low speed limit. This should be a clue that it is invalid.
The discussion is about the usage of appropiate Lorentz transforms. I have shown the appropiate Lorentz transforms for rotating frames, with the associate references from peer - reviewed journals. Would you please let kev finish the derivation?

 

[Dale]

...for \gamma_u=1. A very important detail that I have pointed out to you.

And a detail which I have pointed out to you is wrong. We are going around in circles, which I suppose is appropriate given the subject of the thread. To avoid further circulation and to determine where our disagreement stems could you please identify which of the following you disagree with:

1) In the inertial reference frame where the center is at rest and the mass is moving in the x-y plane along a circle of radius r at an angular speed w the four-acceleration is given by:
\mathbf a=c \frac{d\mathbf u/dt}{|d\mathbf s/dt|} = (0,\; -\gamma^2 r \omega^2 \; cos(\phi + \omega t),\; -\gamma^2 r \omega^2 \; sin(\phi + \omega t),\; 0 )
where
\gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2}

2) The Minkowski norm of this four-acceleration is given by:
|\mathbf a|^2=-\gamma^4 r^2 \omega^4

3) The Minkowski norm of any four-vector is frame invariant meaning that if you calculate it in one frame it will be the same in any other frame whether inertial or non-inertial.

4) The magnitude of the Euclidean norm of the proper acceleration is equal to the magnitude of the Minkowski norm of the four-acceleration.

5) The Euclidean norm of the proper acceleration is given by:
a=\gamma^2 r \omega^2

Look at it this way: your formula produces an infinite acceleration for \omega->\frac{c}{r}. This is unphysical.

No, not only is that not unphysical it is correct. For any finite r, the acceleration measured by an accelerometer goes to infinity in the limit as the tangential velocity goes to c.

Why do you keep diverting the discussion towards the derivation of proper acceleration? The discussion is about the usage of appropiate Lorentz transforms. Would you please let kev finish the derivation?

I'm sure our discussion won't prevent kev from finishing the derivation if he wants to.

 

[yuiop]

If they are correct, then when t' = 0, \sin(\gamma\omega t') = 0 and \cos(\gamma\omega t') = 1 and the equations simplify to:

Incorrect. In the comoving frame you need to realize that \frac{dx'}{dt'}=0 and \frac{dy'}{dt'}=0. Try continuing the calculations from this hint.

I do set \frac{dx'}{dt'}=0 and \frac{dy'}{dt'}=0 in the very next step.

Also \sin(0) = 0 and \cos(0) = 1 are indesputable mathematical facts.

 

[starthaus]

If they are correct, then when t' = 0, \sin(\gamma\omega t') = 0 and \cos(\gamma\omega t') = 1 and the equations simplify to:
[\quote]

I do set \frac{dx'}{dt'}=0 and \frac{dy'}{dt'}=0 in the very next step.

Not in the correct way. You are also setting t'=0 for no reason whatsoever. I have already pointed out this error in the previous post.
New hint: what can you conclude about \frac{d^2x'}{dt'^2} and \frac{d^2y'}{dt'^2}?

Also \sin(0) = 0 and \cos(0) = 1 are indesputable mathematical facts.

True but totally useless for completing the derivation.

 

[yuiop]

P.S. I looked for both of those references. I didn't have access to the first, and I couldn't find the second at all. I can use the usual non-generalized Lorentz transform to answer any questions that can be expressed in terms of a MCIF, but I won't be able to use these non-inertial generalizations.

This is the second reference Starthaus uses: http://arxiv.org/PS_cache/gr-qc/pdf/9904/9904078v2.pdf

 

[starthaus]

Since the vote is split 50/50 I will appoint myself vice president and cast the deciding vote in favor of posting it

So throughout this I will use the convention that the time coordinate is the first coordinate and that timelike intervals squared are positive. So given a wheel centered at the origin and rotating about the z axis in a counter clockwise direction we can describe the worldline for any particle on the rim by:
\mathbf s =(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)

Both Nelson and Nikolic would disagree with the above, I think that this is the origin of your error. Nikolic gives the correct expression for s.
May I suggest that you open a different thread where we can discuss this?

 

[yuiop]

Gron defines a non-inertial metric for a rotating reference frame on Page 89 of his book http://books.google.co.uk/books?id=IyJhCHAryuUC&pg=PA89&lpg=PA89#v=onepage&q&f=false that might be of interest. He uses cylindrical coordinates which seems more natural in this situation, rather than the rectangular coordinates used by Starthaus.

 

[Dale]

we can describe the worldline for any particle on the rim by:
\mathbf s =(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)

Both Nelson and Nikolic would disagree with the above, I think that this is the origin of your error. Nikolic gives the correct expression for s.

So what? I can cite people who agree with me:
http://arxiv.org/PS_cache/hep-th/pdf/0301/0301141v1.pdf
http://physics.bu.edu/~duffy/py105/SHM.html
Plus since I don't have access to those references I cannot check the likely situation that you are misunderstanding what Nelson and Nikolic actually claim.

Instead of an appeal to authority why don't you show what is wrong with it? I mean, the worldline of a particle undergoing uniform circular motion is a circular helix and that is the equation of a helix. So what could possibly be incorrect? Also, according to Nikolic what is the correct expression for the worldline of a particle undergoing uniform circular motion in some inertial frame?

 

[yuiop]

Incorrect. In the comoving frame you need to realize that \frac{dx'}{dt'}=0 and \frac{dy'}{dt'}=0. Try continuing the calculations from this hint.

When I do that, I get:

a_x = \frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\cos(\gamma\omega t')-\gamma\frac{d^2y'}{dt'^2}\sin(\gamma\omega t')- R\gamma^2\omega^2\cos(\gamma\omega t')\right)

a_y = \frac{d^2y}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\sin(\gamma\omega t')+\gamma\frac{d^2y'}{dt'^2}\cos(\gamma\omega t') - R\gamma^2\omega^2\sin(\gamma\omega t')\right)

What are you claiming that tells us?

By the way, if I set the acceleration terms in the rotating frame to zero, the result is:

a_x = \frac{d^2x}{dt^2}=\gamma^{-2}\left(- R\gamma^2\omega^2\cos(\gamma\omega t')\right)

a_y = \frac{d^2y}{dt^2}=\gamma^{-2}\left(- R\gamma^2\omega^2\sin(\gamma\omega t')\right)

When t'=0, the result is:

a_x = \frac{d^2x}{dt^2}= -R\omega^2

a_y = \frac{d^2y}{dt^2}= 0

\sqrt{a_x^2 + a_y^2} = R\omega^2\right)

and when t' is set to a value such that (\gamma\omega t')=\pi/2 (ie a quarter of a turn later) when the x' axis is aligned with the y axis, the result is:

a_x = \frac{d^2x}{dt^2}= 0

a_y = \frac{d^2y}{dt^2}= -R\omega^2

\sqrt{a_x^2 + a_y^2} = R\omega^2

That tells me I can set t' to any value I like and the magnitude of the centripetal acceleration remains constant. Since it is the magnitude we are interested in, in this thread, I am free to set t'=0 for convenience (as I did) and consider that as representative of any arbitary value of t'. This also follows trivially from the circular symmetry as jorrie has been trying to tell you.

P.S. @Dalespam: Did you try the link I gave in #116?

 

[yuiop]

It appears to me as if you have a different definition of comoving frame than the rest of us. We are usually referring to the momentarily comoving inertial frame (MCIF). Or do I read you incorrectly?

The difference between the approach taken by Starthaus compared to the rest of us, is that he is considering the case of a co-moving co-accelerating observer such that this co-accelerating observer measures the acceleration of the particle moving in a circle to be zero.

In the MCIF the co-moving inertial observer is momentarily at rest with the test particle, but it has non-zero acceleration. This is the proper acceleration of the test particle obtained by Dalespam and is what an accelerometer attached to the test particle in the rotating frame would measure.

 

[Dale]

P.S. @Dalespam: Did you try the link I gave in #116?

Yes, my experience working in non-inertial frames is limited, but that looks like a good place to start. If I get a chance to work it out I will post the results here.

 

[starthaus]

When I do that, I get:

a_x = \frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\cos(\gamma\omega t')-\gamma\frac{d^2y'}{dt'^2}\sin(\gamma\omega t')- R\gamma^2\omega^2\cos(\gamma\omega t')\right)

a_y = \frac{d^2y}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\sin(\gamma\omega t')+\gamma\frac{d^2y'}{dt'^2}\cos(\gamma\omega t') - R\gamma^2\omega^2\sin(\gamma\omega t')\right)

What are you claiming that tells us?

By the way, if I set the acceleration terms in the rotating frame to zero, the result is:

a_x = \frac{d^2x}{dt^2}=\gamma^{-2}\left(- R\gamma^2\omega^2\cos(\gamma\omega t')\right)

a_y = \frac{d^2y}{dt^2}=\gamma^{-2}\left(- R\gamma^2\omega^2\sin(\gamma\omega t')\right)

OK, yoy are done. Finally.
All you have to do is to reduce the terms in \gamma^2
The nice thing about this method is that you not only get the magnitude but you also get the components of the acceleration. You get the COMPLETE transforms.

When t'=0, the result is:

a_x = \frac{d^2x}{dt^2}= -R\omega^2

a_y = \frac{d^2y}{dt^2}= 0

Why do you persist in the error of "setting t'=0". You finally got the general transforms, valid for ANY t'.

 

[starthaus]

The difference between the approach taken by Starthaus compared to the rest of us, is that he is considering the case of a co-moving co-accelerating observer such that this co-accelerating observer measures the acceleration of the particle moving in a circle to be zero.

In the MCIF the co-moving inertial observer is momentarily at rest with the test particle, but it has non-zero acceleration. This is the proper acceleration of the test particle obtained by Dalespam and is what an accelerometer attached to the test particle in the rotating frame would measure.

Umm, no. What do you think \frac{d^2x}{dt^2} is?

 

[yuiop]

Umm, no. What do you think \frac{d^2x}{dt^2} is?

That is one component of the COORDINATE centripetal acceleration as measured in the non rotating inertial frame. What do you think it is??

More generally:

The Euclidean norm of the coordinate centripetal acceleration is R\omega^2

This is what is measured in the non rotating inertial frame.

The Euclidean norm of the proper centripetal acceleration is \gamma^2R\omega^2

by the normal definition of proper acceleration, which is equivalent to the four acceleration, or what is measured by an accelerometer.

You have your own unusual definition of proper acceleration as being the acceleration measured by a co-accelerating observer, which always has the magnitude zero.

 

[yuiop]

Why do you persist in the error of "setting t'=0". You finally got the general transforms, valid for ANY t'.

If the transforms are valid for ANY t' then setting t'=0 is NOT an error as you claim. It is just a convenience for working out a particular case that is representitive for magnitude quantities. It so happens that I demonstrated that it is representative for magnitude quantities, by using a diiferent value of t' and getting the same magnitude.

 

[starthaus]

That is one component of the COORDINATE centripetal acceleration as measured in the non rotating inertial frame. What do you think it is??

Good, you finally learned to correct derivation, using the appropiate Lorentz transforms.
So, what does frame S' signify?

 

[starthaus]

If the transforms are valid for ANY t' then setting t'=0 is NOT an error as you claim. It is just a convenience for working out a particular case that is representitive for magnitude quantities. It so happens that I demonstrated that it is representative for magnitude quantities, by using a diiferent value of t' and getting the same magnitude.

Why would you insist on setting t'=0 when I guided you to deriving the general transforms, valid for any t'?
As to the magnitude, you can derive it directly from the general transforms, right? You don't need to insist on making t'=0.

 

[yuiop]

The nice thing about this method is that you not only get the magnitude but you also get the components of the acceleration. You get the COMPLETE transforms.

Yes, it is sometimes useful to have the components and directions and I also like your transform for other reasons. In fact I like it so much that I would to see the derivation steps broken down a bit more so that I confirm the steps you use to obtain it. So far, you have ignored that request, so I can only hope that a better mathematician than myself can verify the steps you take.

However, nothing in your transform disproves anything I said in #1 or anything Dalespam or Jorrie said. We just differ on the definition of proper acceleration.

Unfortunately, you do not seem to understand the physical significance of your own transformations or the four vector quantities that Dalespam is using.

 

[yuiop]

Why do you persist in the error of "setting t'=0". You finally got the general transforms, valid for ANY t'.

Yes, I got the general transforms and then applied them to a specific case. What is wrong with that? That is what general transforms are for.

 

[starthaus]

Gron defines a non-inertial metric for a rotating reference frame on Page 89 of his book http://books.google.co.uk/books?id=IyJhCHAryuUC&pg=PA89&lpg=PA89#v=onepage&q&f=false that might be of interest. He uses cylindrical coordinates which seems more natural in this situation, rather than the rectangular coordinates used by Starthaus.

Thank you, this is a very good reference.
Look at (5.2). What does it tell you?
Specifically, what is the transform for \frac{d^2 \theta}{dt^2}?

 

[starthaus]

Yes, I got the general transforms and then applied them to a specific case. What is wrong with that? That is what general transforms are for.

There is no reason to make t'=0, I have already taught you how to get the general transforms.

 

[yuiop]

Look at (5.2). What does it tell you?
Specifically, what is the transform for \frac{d^2 \theta}{dt^2}?

5.2 says t=T which tells me that Gron is using an unusual clock synchronisation method where the clocks in the rotating frame are articificailly sped up so that stay in sync with the non rotating inertial clock at the centre. Unsual but not invalid if you know what you doing. t is NOT proper time in the normal sense, so you can NOT directly draw any conclusions about proper acceleration by directly using those coordinates. \frac{d^2 \theta}{dt^2} is NOT proper acceleration in the normal sense.

 

[starthaus]

5.2 says t=T which tells me that Gron is using an unusual clock synchronisation method where the clocks in the rotating frame are articificailly sped up so that stay in sync with the non rotating inertial clock at the centre. Unsual but not invalid if you know what you doing. t is NOT proper time in the normal sense, so you can NOT directly draw any conclusions about proper acceleration by directly using those coordinates.

Sure you can. So, you don't like your own reference any more, eh?
This is a pretty good book, pretty recent as well.

\frac{d^2 \theta}{dt^2} is NOT proper acceleration in the normal sense.

Huh?

Anyways, the question was : can you derive its transform between RF and IF? It is not a trick question.

 

[Jorrie]

However, nothing in your transform disproves anything I said in #1 or anything Dalespam or Jorrie said. We just differ on the definition of proper acceleration.

Yes, the differences in definitions (of comoving frame and of proper acceleration) have caused a lot of unnecessary bickering in this thread. I have sympathy with students trying to follow what was going on.

By definition, the centripetal force is always radial, so polar (or cylindrical) coordinates seem to be more appropriate than the Cartesian coordinates that Starthaus used. It reminds me about CarlB's efforts of some years ago to cast the Schwarzschild coordinate accelerations into Cartesian form. He succeeded, but the resultant equations were cumbersome and according to my testing, actually used more fp-calcs than the polar coordinate equations. His motivation was to make orbit simulations faster...

I guess there will be applications where the Cartesian components of centripetal accelerations or forces will be very useful, but the mass swinging at the end of a string is probably not one of them.

 

[starthaus]

Yes, the differences in definitions (of comoving frame and of proper acceleration) have caused a lot of unnecessary bickering in this thread. I have sympathy with students trying to follow what was going on.

By definition, the centripetal force is always radial, so polar (or cylindrical) coordinates seem to be more appropriate than the Cartesian coordinates that Starthaus used.

It is not about that, it is about teaching kev how to use the appropiate transforms and how to derive the correct value for the proper acceleration. His OP was wong on both accounts. It took 9 pages and 100+ posts to get the correct results for what should have taken two iterations of derivatives.

 

[Jorrie]

It is not about that, it is about teaching kev how to use the appropiate transforms and how to derive the correct value for the proper acceleration. His OP was wong on both accounts.

No, I think his OP is perfectly valid, by his (and AFAIK, all around here, but yourself) definition of proper acceleration - measured as he stated, by a tension gauge oriented radially (implied by 'centripetal force'). He has not shown a rigorous derivation, but that has been done before by pervect, amongst others.

It took 9 pages and 100+ posts to get the correct results for what should have taken two iterations of derivatives.

I do not quite see how the equations that kev posted in https://www.physicsforums.com/showpost.php?p=2693883&postcount=120" and which you said are correct, produce what the tension gauge (or an accelerometer) measures on the circling mass. Can you please enlighten us on that?

 

[starthaus]

I do not quite see how the equations that kev posted in https://www.physicsforums.com/showpost.php?p=2693883&postcount=120" and which you said are correct, produce what the tension gauge (or an accelerometer) measures on the circling mass. Can you please enlighten us on that?

Give it some time, it will sink in.

 

[Dale]

Hi starthaus, if you think that the value you have had kev derive in #120 represents the proper acceleration of the particle then you are simply wrong.

 

[starthaus]

Hi starthaus, if you think that the value you have had kev derive in #120 represents the proper acceleration of the particle then you are simply wrong.

I have guided kev in doing the correct derivation of the acceleration measured in the inertial frame S as a function of the coordinates in the rotating frame S' by using the appropiate Lorentz transforms. This closes the argument started at post #3. Nothing more , or less.

 

[Dale]

That is correct, it is the coordinate acceleration in the original inertial frame transformed to the rotating frame. It is not the proper acceleration.

 

[starthaus]

That is correct, it is the coordinate acceleration in the original inertial frame transformed to the rotating frame.

Thank you.

It is not the proper acceleration.

May I suggest that you open a different thread that discusses how to obtain the proper acceleration from the correct line element (see Gron's book)? The line element "s" that you used in your derivation is incorrect.

 

[Dale]

May I suggest that you open a different thread that discusses how to obtain the proper acceleration from the correct line element (see Gron's book)? The line element "s" that you used in your derivation is incorrect.

That equation was not the line element, it was the worldline of the particle. A line element is a scalar, the worldline is a four-vector parameterized by some arbitrary scalar.

 

[starthaus]

That equation was not the line element, it was the worldline of the particle. A line element is a scalar, the worldline is a four-vector parameterized by some arbitrary scalar.

You volunteered to study the chapter on rotating frames in Gron's book. Why don't we reprise this discussion in a different thread, once you have studied the chapter?

 

[Dale]

You volunteered to study the chapter on rotating frames in Gron's book. Why don't we reprise this discussion in a different thread, once you have studied the chapter?

Do you or do you not agree that the worldline of a particle undergoing uniform circular motion in some inertial reference frame is given by:
(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)

 

[starthaus]

Do you or do you not agree that the worldline of a particle undergoing uniform circular motion in some inertial reference frame is given by:
(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)

Why don't you spend some time understanding the Gron chapter on the subject? Once you do that, we can talk in a separate thread about how to calculate the proper acceleration. The value you calculated is wrong, there is no point in muddling this thread.

 

[Dale]

You are really starting to irritate me with your repeated assertions that it is wrong followed by a completely evasive non-answer every time your assertion is challenged. You have stated in this thread that my derivation is wrong so defend your statement in this thread and stop trying to weasel your way out of it.

If that expression does not represent the worldline of a particle undergoing uniform circular motion in some inertial reference frame then what expression does?

 

[starthaus]

You are really starting to irritate me with your repeated assertions that it is wrong followed by a completely evasive non-answer every time your assertion is challenged. You have stated in this thread that my derivation is wrong so defend your statement in this thread and stop trying to weasel your way out of it.

Since you don't want to read the Gron chapter, I'll help you out:

The transformation between IF and RF is:

t=T
\theta=\Theta-\omega T

If you calculate the proper acceleration \frac{d^2\Theta}{dT^2} you find out that it is equal to the coordinate acceleration in IF \frac{d^2\theta}{dt^2} . Earlier in this thread (post 120), the coordinate acceleration in IF has been found to be R\omega^2. You can read the same exact value straight of Gron's line element. What does this tell you?

If that expression does not represent the worldline of a particle undergoing uniform circular motion in some inertial reference frame then what expression does?

The answer is found in the Nikolic paper. I provided a link to it several times.

 

[Dale]

OK, so according to Gron the worldline of a particle undergoing uniform circular motion:
In the rotating frame is given by:
(ct,r_0,\theta_0,0)

And in an inertial frame using cylindrical coordinates is given by:
(cT,r_0,\theta_0+\omega T,0)

Transforming this into the inertial frame using Cartesian coordinates we obtain:
(ct,\; r_0 \; cos(\theta_0 + \omega t),\; r_0 \; sin(\theta_0 + \omega t),\; 0)

Which is the same form as:
(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)

So Gron agrees with me wrt the form of the worldline.

 

[starthaus]

So Gron agrees with me wrt the form of the worldline.

...but not on your derivation for the acceleration. This is very simple stuff, why is so difficult for you to admit that you are wrong?